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Question Number 76431 by john santu last updated on 27/Dec/19
how to find x^(2048)  + x^(−2048)   if  x+x^(−1) =(((√5)+1)/2)
howtofindx2048+x2048ifx+x1=5+12
Commented by mathmax by abdo last updated on 28/Dec/19
we have x+x^(−1)  =((1+(√5))/2) ⇒(x+x^(−1) )^n  =(((1+(√5))/2))^n  ⇒  Σ_(k=0) ^n  C_n ^k  x^k  (x^(−1) )^(n−k)  =(((1+(√5))/2))^n  ⇒  Σ_(k=0) ^n  C_n ^k  x^k  x^(k−n)  =(((1+(√5))/2))^n   ⇒Σ_(k=0) ^n  C_n ^k  x^(2k)  =(((1+(√5))/2))^n  x^n   change x by (1/x) ⇒Σ_(k=0) ^n  x^(−2k)  =(((1+(√5))/2))^(n ) x^(−n)  ⇒  Σ_(k=0) ^n  C_n ^k x^(2k)  +Σ_(k=0) ^n  C_n ^k  x^(−2k)  =(((1+(√5))/2))^n (x^n  +x^(−n) ) ⇒  x^n  +x^(−n ) =((Σ_(k=0) ^n  C_n ^k  x^(2k)  +Σ_(k=0) ^n  C_n ^k  x^(−2k) )/((((1+(√5))/2))^n )) =(((x^2 +1)^n  +(x^(−2)  +1)^n )/((((1+(√5))/2))^n )) ⇒  x^(2048)  +x^(−2048)   =(((1+x^2 )^(2048)  +(1+x^(−2) )^(2048) )/((((1+(√5))/2))^(2048) ))
wehavex+x1=1+52(x+x1)n=(1+52)nk=0nCnkxk(x1)nk=(1+52)nk=0nCnkxkxkn=(1+52)nk=0nCnkx2k=(1+52)nxnchangexby1xk=0nx2k=(1+52)nxnk=0nCnkx2k+k=0nCnkx2k=(1+52)n(xn+xn)xn+xn=k=0nCnkx2k+k=0nCnkx2k(1+52)n=(x2+1)n+(x2+1)n(1+52)nx2048+x2048=(1+x2)2048+(1+x2)2048(1+52)2048
Answered by Kunal12588 last updated on 27/Dec/19
x+x^(−1) =(((√5)+1)/2)  x^2 +2+x^(−2) =((6+2(√5))/4)=((3+(√5))/2)  x^2 +x^(−2) =(((√5)−1)/2)  x^4 +x^(−4) =((6−2(√5))/4)−2=−((1+(√5))/2)  x^8 +x^(−8) =(((√5)−1)/2)  x^(16) +x^(−16) =−((1+(√5))/2)  ⋮  x^(2048) +x^(−2048) =(((√5)−1)/2)
x+x1=5+12x2+2+x2=6+254=3+52x2+x2=512x4+x4=62542=1+52x8+x8=512x16+x16=1+52x2048+x2048=512
Commented by john santu last updated on 27/Dec/19
thanks sir
thankssir
Commented by mr W last updated on 27/Dec/19
what if it is to find x^(2020) +x^(−2020)  ?
whatifitistofindx2020+x2020?
Commented by Kunal12588 last updated on 27/Dec/19
I can find x  x=(((1+(√5))±i(√(2(√5)((√5)−1))))/4)  but how to get x^(2020)  and  (1/x^(2020) )?
Icanfindxx=(1+5)±i25(51)4buthowtogetx2020and1x2020?
Commented by Kunal12588 last updated on 27/Dec/19
f_n =x^n +x^(−n)   given : f_1 =(((√5)+1)/2)  find:  f_(2048) ,f_(2020)  and f_n
fn=xn+xngiven:f1=5+12find:f2048,f2020andfn

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